What is the equation of the line normal to $f(x)=-sqrt((x-1)(x+3)$ at $x=0$ ?

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Answer 1 · 2015-11-26T02:59:13

f(x) does not exist at x=0, hence normal to f(x) does not exist there.

At x=0, f(x) is $-sqrt(-3)$ , which is an imaginary number hence it does not exist.

What is the equation of the line normal to f(x)=-sqrt((x-1)(x+3) f(x)=-sqrt((x-1)(x+3) at x=0 x=0?